question_id int64 1 75.7k | db_id stringclasses 33
values | db_name stringclasses 4
values | question stringlengths 19 259 | partition stringclasses 4
values | difficulty stringclasses 3
values | SQL stringlengths 25 862 |
|---|---|---|---|---|---|---|
1,001 | restaurant_bills | spider | Show distinct managers of branches. | test | easy | SELECT DISTINCT manager FROM branch |
1,002 | restaurant_bills | spider | Who are the distinct managers of branches? | test | easy | SELECT DISTINCT manager FROM branch |
1,003 | restaurant_bills | spider | List the names of customers that do not have any order. | test | medium | SELECT name FROM customer WHERE NOT customer_id IN (SELECT customer_id FROM customer_order) |
1,004 | restaurant_bills | spider | Which customers do not have any order? Give me the customer names. | test | medium | SELECT name FROM customer WHERE NOT customer_id IN (SELECT customer_id FROM customer_order) |
1,005 | real_estate_rentals | spider | Which countries and cities are included in addresses? | test | easy | SELECT country, town_city FROM addresses |
1,006 | real_estate_rentals | spider | What are the countries and cities for each address? | test | easy | SELECT country, town_city FROM addresses |
1,007 | real_estate_rentals | spider | In which states are each of the the properties located? | test | hard | SELECT DISTINCT t1.county_state_province FROM addresses AS t1 JOIN properties AS t2 ON t1.address_id = t2.property_address_id |
1,008 | real_estate_rentals | spider | Give the states or provinces corresponding to each property. | test | hard | SELECT DISTINCT t1.county_state_province FROM addresses AS t1 JOIN properties AS t2 ON t1.address_id = t2.property_address_id |
1,009 | real_estate_rentals | spider | How is the feature rooftop described? | test | easy | SELECT feature_description FROM features WHERE feature_name = 'rooftop' |
1,010 | real_estate_rentals | spider | Return the description of the feature 'rooftop'. | test | easy | SELECT feature_description FROM features WHERE feature_name = 'rooftop' |
1,011 | real_estate_rentals | spider | What are the feature name and description of the most commonly seen feature across properties? | test | hard | SELECT t1.feature_name, t1.feature_description FROM features AS t1 JOIN property_features AS t2 ON t1.feature_id = t2.feature_id GROUP BY t1.feature_name ORDER BY COUNT(*) DESC LIMIT 1 |
1,012 | real_estate_rentals | spider | Give the feature name and description for the most common feature across all properties. | test | hard | SELECT t1.feature_name, t1.feature_description FROM features AS t1 JOIN property_features AS t2 ON t1.feature_id = t2.feature_id GROUP BY t1.feature_name ORDER BY COUNT(*) DESC LIMIT 1 |
1,013 | real_estate_rentals | spider | What is the minimum number of rooms in a property? | test | easy | SELECT MIN(room_count) FROM properties |
1,014 | real_estate_rentals | spider | What is the lowest room count across all the properties? | test | easy | SELECT MIN(room_count) FROM properties |
1,015 | real_estate_rentals | spider | How many properties have 1 parking lot or 1 garage? | test | medium | SELECT COUNT(*) FROM properties WHERE parking_lots = 1 OR garage_yn = 1 |
1,016 | real_estate_rentals | spider | Count the number of properties that have 1 parking lot or 1 garage. | test | medium | SELECT COUNT(*) FROM properties WHERE parking_lots = 1 OR garage_yn = 1 |
1,017 | real_estate_rentals | spider | For users whose description contain the string 'Mother', which age categories are they in? | test | hard | SELECT t2.age_category_code FROM ref_user_categories AS t1 JOIN users AS t2 ON t1.user_category_code = t2.user_category_code WHERE t1.user_category_description LIKE '%mother' |
1,018 | real_estate_rentals | spider | What are the age categories for users whose description contains the string Mother? | test | hard | SELECT t2.age_category_code FROM ref_user_categories AS t1 JOIN users AS t2 ON t1.user_category_code = t2.user_category_code WHERE t1.user_category_description LIKE '%mother' |
1,019 | real_estate_rentals | spider | What is the first name of the user who owns the greatest number of properties? | test | hard | SELECT t1.first_name FROM users AS t1 JOIN properties AS t2 ON t2.owner_user_id = t1.user_id GROUP BY t1.user_id ORDER BY COUNT(*) DESC LIMIT 1 |
1,020 | real_estate_rentals | spider | Return the first name of the user who owns the most properties. | test | hard | SELECT t1.first_name FROM users AS t1 JOIN properties AS t2 ON t2.owner_user_id = t1.user_id GROUP BY t1.user_id ORDER BY COUNT(*) DESC LIMIT 1 |
1,021 | real_estate_rentals | spider | List the average room count of the properties with gardens. | test | hard | SELECT AVG(t3.room_count) FROM property_features AS t1 JOIN features AS t2 ON t1.feature_id = t2.feature_id JOIN properties AS t3 ON t1.property_id = t3.property_id WHERE t2.feature_name = 'garden' |
1,022 | real_estate_rentals | spider | On average, how many rooms do properties with garden features have? | test | hard | SELECT AVG(t3.room_count) FROM property_features AS t1 JOIN features AS t2 ON t1.feature_id = t2.feature_id JOIN properties AS t3 ON t1.property_id = t3.property_id WHERE t2.feature_name = 'garden' |
1,023 | real_estate_rentals | spider | In which cities are there any properties equipped with a swimming pool? | test | hard | SELECT t2.town_city FROM properties AS t1 JOIN addresses AS t2 ON t1.property_address_id = t2.address_id JOIN property_features AS t3 ON t1.property_id = t3.property_id JOIN features AS t4 ON t4.feature_id = t3.feature_id WHERE t4.feature_name = 'swimming pool' |
1,024 | real_estate_rentals | spider | Return the cities in which there exist properties that have swimming pools. | test | hard | SELECT t2.town_city FROM properties AS t1 JOIN addresses AS t2 ON t1.property_address_id = t2.address_id JOIN property_features AS t3 ON t1.property_id = t3.property_id JOIN features AS t4 ON t4.feature_id = t3.feature_id WHERE t4.feature_name = 'swimming pool' |
1,025 | real_estate_rentals | spider | Which property had the lowest price requested by the vendor? List the id and the price. | test | hard | SELECT property_id, vendor_requested_price FROM properties ORDER BY vendor_requested_price LIMIT 1 |
1,026 | real_estate_rentals | spider | What is the id of the property that had the lowest requested price from the vendor, and what was that price? | test | hard | SELECT property_id, vendor_requested_price FROM properties ORDER BY vendor_requested_price LIMIT 1 |
1,027 | real_estate_rentals | spider | On average, how many rooms does a property have? | test | easy | SELECT AVG(room_count) FROM properties |
1,028 | real_estate_rentals | spider | What is the average number of rooms in a property? | test | easy | SELECT AVG(room_count) FROM properties |
1,029 | real_estate_rentals | spider | How many kinds of room sizes are listed? | test | easy | SELECT COUNT(DISTINCT room_size) FROM rooms |
1,030 | real_estate_rentals | spider | Return the number of different room sizes. | test | easy | SELECT COUNT(DISTINCT room_size) FROM rooms |
1,031 | real_estate_rentals | spider | What are the ids of users who have searched at least twice, and what did they search? | test | hard | SELECT search_seq, user_id FROM user_searches GROUP BY user_id HAVING COUNT(*) >= 2 |
1,032 | real_estate_rentals | spider | Return the ids of users who have performed two or more searches, as well as their search sequence. | test | hard | SELECT search_seq, user_id FROM user_searches GROUP BY user_id HAVING COUNT(*) >= 2 |
1,033 | real_estate_rentals | spider | When was the time of the latest search by a user? | test | easy | SELECT MAX(search_datetime) FROM user_searches |
1,034 | real_estate_rentals | spider | What was the time of the most recent search? | test | easy | SELECT MAX(search_datetime) FROM user_searches |
1,035 | real_estate_rentals | spider | What are all the user searches time and content? Sort the result descending by content. | test | hard | SELECT search_datetime, search_string FROM user_searches ORDER BY search_string DESC |
1,036 | real_estate_rentals | spider | Return the search strings and corresonding time stamps for all user searches, sorted by search string descending. | test | hard | SELECT search_datetime, search_string FROM user_searches ORDER BY search_string DESC |
1,037 | real_estate_rentals | spider | What are the zip codes of properties which do not belong to users who own at most 2 properties? | test | hard | SELECT t1.zip_postcode FROM addresses AS t1 JOIN properties AS t2 ON t1.address_id = t2.property_address_id WHERE NOT t2.owner_user_id IN (SELECT owner_user_id FROM properties GROUP BY owner_user_id HAVING COUNT(*) <= 2) |
1,038 | real_estate_rentals | spider | Return the zip codes for properties not belonging to users who own two or fewer properties. | test | hard | SELECT t1.zip_postcode FROM addresses AS t1 JOIN properties AS t2 ON t1.address_id = t2.property_address_id WHERE NOT t2.owner_user_id IN (SELECT owner_user_id FROM properties GROUP BY owner_user_id HAVING COUNT(*) <= 2) |
1,039 | real_estate_rentals | spider | What are the users making only one search? List both category and user id. | test | hard | SELECT t1.user_category_code, t1.user_id FROM users AS t1 JOIN user_searches AS t2 ON t1.user_id = t2.user_id GROUP BY t1.user_id HAVING COUNT(*) = 1 |
1,040 | real_estate_rentals | spider | What are the ids of users who have only made one search, and what are their category codes? | test | hard | SELECT t1.user_category_code, t1.user_id FROM users AS t1 JOIN user_searches AS t2 ON t1.user_id = t2.user_id GROUP BY t1.user_id HAVING COUNT(*) = 1 |
1,041 | real_estate_rentals | spider | What is the age range category of the user who made the first search? | test | hard | SELECT t1.age_category_code FROM users AS t1 JOIN user_searches AS t2 ON t1.user_id = t2.user_id ORDER BY t2.search_datetime LIMIT 1 |
1,042 | real_estate_rentals | spider | Return the age category for the user who made the earliest search. | test | hard | SELECT t1.age_category_code FROM users AS t1 JOIN user_searches AS t2 ON t1.user_id = t2.user_id ORDER BY t2.search_datetime LIMIT 1 |
1,043 | real_estate_rentals | spider | Find the login names of all senior citizen users ordered by their first names. | test | medium | SELECT login_name FROM users WHERE user_category_code = 'senior citizen' ORDER BY first_name |
1,044 | real_estate_rentals | spider | What are the login names of all senior citizens, sorted by first name? | test | medium | SELECT login_name FROM users WHERE user_category_code = 'senior citizen' ORDER BY first_name |
1,045 | real_estate_rentals | spider | How many searches do buyers make in total? | test | hard | SELECT COUNT(*) FROM users AS t1 JOIN user_searches AS t2 ON t1.user_id = t2.user_id WHERE t1.is_buyer = 1 |
1,046 | real_estate_rentals | spider | Count the number of searches made by buyers. | test | hard | SELECT COUNT(*) FROM users AS t1 JOIN user_searches AS t2 ON t1.user_id = t2.user_id WHERE t1.is_buyer = 1 |
1,047 | real_estate_rentals | spider | When did the user with login name ratione register? | test | easy | SELECT date_registered FROM users WHERE login_name = 'ratione' |
1,048 | real_estate_rentals | spider | What was the registration date for the user whose login name is ratione? | test | easy | SELECT date_registered FROM users WHERE login_name = 'ratione' |
1,049 | real_estate_rentals | spider | List the first name, middle name and last name, and log in name of all the seller users, whose seller value is 1. | test | easy | SELECT first_name, middle_name, last_name, login_name FROM users WHERE is_seller = 1 |
1,050 | real_estate_rentals | spider | What are the first, middle, last, and login names for all users who are sellers? | test | easy | SELECT first_name, middle_name, last_name, login_name FROM users WHERE is_seller = 1 |
1,051 | real_estate_rentals | spider | Where do the Senior Citizens live? List building, street, and the city. | test | hard | SELECT t1.line_1_number_building, t1.line_2_number_street, t1.town_city FROM addresses AS t1 JOIN users AS t2 ON t1.address_id = t2.user_address_id WHERE t2.user_category_code = 'senior citizen' |
1,052 | real_estate_rentals | spider | What are the buildings, streets, and cities corresponding to the addresses of senior citizens? | test | hard | SELECT t1.line_1_number_building, t1.line_2_number_street, t1.town_city FROM addresses AS t1 JOIN users AS t2 ON t1.address_id = t2.user_address_id WHERE t2.user_category_code = 'senior citizen' |
1,053 | real_estate_rentals | spider | How many properties are there with at least 2 features? | test | hard | SELECT COUNT(*) FROM properties GROUP BY property_id HAVING COUNT(*) >= 2 |
1,054 | real_estate_rentals | spider | Count the number of properties with at least two features. | test | hard | SELECT COUNT(*) FROM properties GROUP BY property_id HAVING COUNT(*) >= 2 |
1,055 | real_estate_rentals | spider | How many photos does each property have? | test | hard | SELECT COUNT(*), property_id FROM property_photos GROUP BY property_id |
1,056 | real_estate_rentals | spider | Count the number of property photos each property has by id. | test | hard | SELECT COUNT(*), property_id FROM property_photos GROUP BY property_id |
1,057 | real_estate_rentals | spider | How many photos does each owner has of his or her properties? List user id and number of photos. | test | hard | SELECT t1.owner_user_id, COUNT(*) FROM properties AS t1 JOIN property_photos AS t2 ON t1.property_id = t2.property_id GROUP BY t1.owner_user_id |
1,058 | real_estate_rentals | spider | What are the user ids of property owners who have property photos, and how many do each of them have? | test | hard | SELECT t1.owner_user_id, COUNT(*) FROM properties AS t1 JOIN property_photos AS t2 ON t1.property_id = t2.property_id GROUP BY t1.owner_user_id |
1,059 | real_estate_rentals | spider | What is the total max price of the properties owned by single mothers or students? | test | hard | SELECT SUM(t1.price_max) FROM properties AS t1 JOIN users AS t2 ON t1.owner_user_id = t2.user_id WHERE t2.user_category_code = 'single mother' OR t2.user_category_code = 'student' |
1,060 | real_estate_rentals | spider | Give the total max price corresponding to any properties owned by single mothers or students. | test | hard | SELECT SUM(t1.price_max) FROM properties AS t1 JOIN users AS t2 ON t1.owner_user_id = t2.user_id WHERE t2.user_category_code = 'single mother' OR t2.user_category_code = 'student' |
1,061 | real_estate_rentals | spider | What are the date stamps and property names for each item of property history, ordered by date stamp? | test | hard | SELECT t1.datestamp, t2.property_name FROM user_property_history AS t1 JOIN properties AS t2 ON t1.property_id = t2.property_id ORDER BY datestamp |
1,062 | real_estate_rentals | spider | Return the date stamp and property name for each property history event, sorted by date stamp. | test | hard | SELECT t1.datestamp, t2.property_name FROM user_property_history AS t1 JOIN properties AS t2 ON t1.property_id = t2.property_id ORDER BY datestamp |
1,063 | real_estate_rentals | spider | What is the description of the most common property type? List the description and code. | test | hard | SELECT t1.property_type_description, t1.property_type_code FROM ref_property_types AS t1 JOIN properties AS t2 ON t1.property_type_code = t2.property_type_code GROUP BY t1.property_type_code ORDER BY COUNT(*) DESC LIMIT 1 |
1,064 | real_estate_rentals | spider | What is the most common property type, and what is its description. | test | hard | SELECT t1.property_type_description, t1.property_type_code FROM ref_property_types AS t1 JOIN properties AS t2 ON t1.property_type_code = t2.property_type_code GROUP BY t1.property_type_code ORDER BY COUNT(*) DESC LIMIT 1 |
1,065 | real_estate_rentals | spider | What is the detailed description of the age category code 'Over 60'? | test | medium | SELECT age_category_description FROM ref_age_categories WHERE age_category_code = 'over 60' |
1,066 | real_estate_rentals | spider | Give the category description of the age category 'Over 60'. | test | medium | SELECT age_category_description FROM ref_age_categories WHERE age_category_code = 'over 60' |
1,067 | real_estate_rentals | spider | What are the different room sizes, and how many of each are there? | test | hard | SELECT room_size, COUNT(*) FROM rooms GROUP BY room_size |
1,068 | real_estate_rentals | spider | Return the number of rooms with each different room size. | test | hard | SELECT room_size, COUNT(*) FROM rooms GROUP BY room_size |
1,069 | real_estate_rentals | spider | In which country does the user with first name Robbie live? | test | hard | SELECT t1.country FROM addresses AS t1 JOIN users AS t2 ON t1.address_id = t2.user_address_id WHERE t2.first_name = 'robbie' |
1,070 | real_estate_rentals | spider | Return the country in which the user with first name Robbie lives. | test | hard | SELECT t1.country FROM addresses AS t1 JOIN users AS t2 ON t1.address_id = t2.user_address_id WHERE t2.first_name = 'robbie' |
1,071 | real_estate_rentals | spider | What are the first, middle and last names of users who own the property they live in? | test | hard | SELECT first_name, middle_name, last_name FROM properties AS t1 JOIN users AS t2 ON t1.owner_user_id = t2.user_id WHERE t1.property_address_id = t2.user_address_id |
1,072 | real_estate_rentals | spider | Return the full names of users who live in properties that they own. | test | hard | SELECT first_name, middle_name, last_name FROM properties AS t1 JOIN users AS t2 ON t1.owner_user_id = t2.user_id WHERE t1.property_address_id = t2.user_address_id |
1,073 | real_estate_rentals | spider | List the search content of the users who do not own a single property. | test | hard | SELECT search_string FROM user_searches EXCEPT SELECT t1.search_string FROM user_searches AS t1 JOIN properties AS t2 ON t1.user_id = t2.owner_user_id |
1,074 | real_estate_rentals | spider | What search strings were entered by users who do not own any properties? | test | hard | SELECT search_string FROM user_searches EXCEPT SELECT t1.search_string FROM user_searches AS t1 JOIN properties AS t2 ON t1.user_id = t2.owner_user_id |
1,075 | real_estate_rentals | spider | List the last names and ids of users who have at least 2 properties and searched at most twice. | test | hard | SELECT t1.last_name, t1.user_id FROM users AS t1 JOIN user_searches AS t2 ON t1.user_id = t2.user_id GROUP BY t1.user_id HAVING COUNT(*) <= 2 INTERSECT SELECT t3.last_name, t3.user_id FROM users AS t3 JOIN properties AS t4 ON t3.user_id = t4.owner_user_id GROUP BY t3.user_id HAVING COUNT(*) >= 2 |
1,076 | real_estate_rentals | spider | What are the last names and ids of users who have searched two or fewer times, and own two or more properties? | test | hard | SELECT t1.last_name, t1.user_id FROM users AS t1 JOIN user_searches AS t2 ON t1.user_id = t2.user_id GROUP BY t1.user_id HAVING COUNT(*) <= 2 INTERSECT SELECT t3.last_name, t3.user_id FROM users AS t3 JOIN properties AS t4 ON t3.user_id = t4.owner_user_id GROUP BY t3.user_id HAVING COUNT(*) >= 2 |
1,077 | customers_card_transactions | spider | How many accounts do we have? | train | easy | SELECT COUNT(*) FROM accounts |
1,078 | customers_card_transactions | spider | Count the number of accounts. | train | easy | SELECT COUNT(*) FROM accounts |
1,079 | customers_card_transactions | spider | Show ids, customer ids, names for all accounts. | train | easy | SELECT account_id, customer_id, account_name FROM accounts |
1,080 | customers_card_transactions | spider | What are the account ids, customer ids, and account names for all the accounts? | train | easy | SELECT account_id, customer_id, account_name FROM accounts |
1,081 | customers_card_transactions | spider | Show other account details for account with name 338. | train | easy | SELECT other_account_details FROM accounts WHERE account_name = '338' |
1,082 | customers_card_transactions | spider | What are the other account details for the account with the name 338? | train | easy | SELECT other_account_details FROM accounts WHERE account_name = '338' |
1,083 | customers_card_transactions | spider | What is the first name, last name, and phone of the customer with account name 162? | train | hard | SELECT t2.customer_first_name, t2.customer_last_name, t2.customer_phone FROM accounts AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t1.account_name = '162' |
1,084 | customers_card_transactions | spider | Give the full name and phone of the customer who has the account name 162. | train | hard | SELECT t2.customer_first_name, t2.customer_last_name, t2.customer_phone FROM accounts AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t1.account_name = '162' |
1,085 | customers_card_transactions | spider | How many accounts does the customer with first name Art and last name Turcotte have? | train | hard | SELECT COUNT(*) FROM accounts AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t2.customer_first_name = 'art' AND t2.customer_last_name = 'turcotte' |
1,086 | customers_card_transactions | spider | Return the number of accounts that the customer with the first name Art and last name Turcotte has. | train | hard | SELECT COUNT(*) FROM accounts AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t2.customer_first_name = 'art' AND t2.customer_last_name = 'turcotte' |
1,087 | customers_card_transactions | spider | Show all customer ids and the number of accounts for each customer. | train | hard | SELECT customer_id, COUNT(*) FROM accounts GROUP BY customer_id |
1,088 | customers_card_transactions | spider | How many accounts are there for each customer id? | train | hard | SELECT customer_id, COUNT(*) FROM accounts GROUP BY customer_id |
1,089 | customers_card_transactions | spider | Show the customer id and number of accounts with most accounts. | train | hard | SELECT customer_id, COUNT(*) FROM accounts GROUP BY customer_id ORDER BY COUNT(*) DESC LIMIT 1 |
1,090 | customers_card_transactions | spider | What is the customer id of the customer with the most accounts, and how many accounts does this person have? | train | hard | SELECT customer_id, COUNT(*) FROM accounts GROUP BY customer_id ORDER BY COUNT(*) DESC LIMIT 1 |
1,091 | customers_card_transactions | spider | What is the customer first, last name and id with least number of accounts. | train | hard | SELECT t2.customer_first_name, t2.customer_last_name, t1.customer_id FROM accounts AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id GROUP BY t1.customer_id ORDER BY COUNT(*) ASC LIMIT 1 |
1,092 | customers_card_transactions | spider | Give the full name and customer id of the customer with the fewest accounts. | train | hard | SELECT t2.customer_first_name, t2.customer_last_name, t1.customer_id FROM accounts AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id GROUP BY t1.customer_id ORDER BY COUNT(*) ASC LIMIT 1 |
1,093 | customers_card_transactions | spider | Show the number of all customers without an account. | train | easy | SELECT COUNT(*) FROM customers WHERE NOT customer_id IN (SELECT customer_id FROM accounts) |
1,094 | customers_card_transactions | spider | How many customers do not have an account? | train | easy | SELECT COUNT(*) FROM customers WHERE NOT customer_id IN (SELECT customer_id FROM accounts) |
1,095 | customers_card_transactions | spider | Show the first names and last names of customers without any account. | train | hard | SELECT customer_first_name, customer_last_name FROM customers EXCEPT SELECT t1.customer_first_name, t1.customer_last_name FROM customers AS t1 JOIN accounts AS t2 ON t1.customer_id = t2.customer_id |
1,096 | customers_card_transactions | spider | What are the full names of customers who do not have any accounts? | train | hard | SELECT customer_first_name, customer_last_name FROM customers EXCEPT SELECT t1.customer_first_name, t1.customer_last_name FROM customers AS t1 JOIN accounts AS t2 ON t1.customer_id = t2.customer_id |
1,097 | customers_card_transactions | spider | Show distinct first and last names for all customers with an account. | train | hard | SELECT DISTINCT t1.customer_first_name, t1.customer_last_name FROM customers AS t1 JOIN accounts AS t2 ON t1.customer_id = t2.customer_id |
1,098 | customers_card_transactions | spider | What are the full names of customers who have accounts? | train | hard | SELECT DISTINCT t1.customer_first_name, t1.customer_last_name FROM customers AS t1 JOIN accounts AS t2 ON t1.customer_id = t2.customer_id |
1,099 | customers_card_transactions | spider | How many customers have an account? | train | easy | SELECT COUNT(DISTINCT customer_id) FROM accounts |
1,100 | customers_card_transactions | spider | Count the number of customers who hold an account. | train | easy | SELECT COUNT(DISTINCT customer_id) FROM accounts |
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